JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
Let the tangent to the parabola \(y^2=12 x\) at the point \((3, \alpha)\) be perpendicular to the line \(2 x+2 y=3\).Then the square of distance of the point \((6,-4)\)from the normal to the hyperbola \(\alpha^2 x^2-9 y^2=9 \alpha^2\)at its point \((\alpha-1, \alpha+2)\) is equal to \(........\).
- A \(116\)
- B \(115\)
- C \(114\)
- D \(113\)
Answer & Solution
Correct Answer
(A) \(116\)
Step-by-step Solution
Detailed explanation
\(\because P (3, \alpha)\) lies on \(y^2=12 x\) \(\Rightarrow \alpha= \pm 6\) But, \(\left.\frac{ dy }{ dx }\right|_{(3, \alpha)}=\frac{6}{\alpha}=1 \Rightarrow \alpha=6(\alpha=-6\) reject \()\) Now, hyperbola \(\frac{x^2}{9}-\frac{y^2}{36}=1\), normal at…
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